3.49.24 \(\int \frac {2 e^x x+(4 x^2+e^x (-4 x+2 x^2)) \log (\frac {x}{2})+(4 x^2-4 x^2 \log (\frac {x}{2})) \log (x)+(-e^x+(e^x (1-x)-4 x) \log (\frac {x}{2})+(-4 x+4 x \log (\frac {x}{2})) \log (x)) \log (\log (\log (4)))+(\log (\frac {x}{2})+(1-\log (\frac {x}{2})) \log (x)) \log ^2(\log (\log (4)))}{4 x^4-4 x^3 \log (\log (\log (4)))+x^2 \log ^2(\log (\log (4)))} \, dx\)

Optimal. Leaf size=29 \[ \frac {\log \left (\frac {x}{2}\right ) \left (\log (x)+\frac {e^x}{2 x-\log (\log (\log (4)))}\right )}{x} \]

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Rubi [B]  time = 3.88, antiderivative size = 62, normalized size of antiderivative = 2.14, number of steps used = 26, number of rules used = 11, integrand size = 155, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1594, 27, 6688, 6742, 14, 2304, 2303, 2366, 2178, 2177, 2554} \begin {gather*} \frac {\log (x) \log \left (\frac {x}{2}\right )}{x}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{\log (\log (\log (4))) (2 x-\log (\log (\log (4))))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^x*x + (4*x^2 + E^x*(-4*x + 2*x^2))*Log[x/2] + (4*x^2 - 4*x^2*Log[x/2])*Log[x] + (-E^x + (E^x*(1 - x)
- 4*x)*Log[x/2] + (-4*x + 4*x*Log[x/2])*Log[x])*Log[Log[Log[4]]] + (Log[x/2] + (1 - Log[x/2])*Log[x])*Log[Log[
Log[4]]]^2)/(4*x^4 - 4*x^3*Log[Log[Log[4]]] + x^2*Log[Log[Log[4]]]^2),x]

[Out]

(Log[x/2]*Log[x])/x - (E^x*Log[x/2])/(x*Log[Log[Log[4]]]) + (2*E^x*Log[x/2])/((2*x - Log[Log[Log[4]]])*Log[Log
[Log[4]]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^x x+\left (4 x^2+e^x \left (-4 x+2 x^2\right )\right ) \log \left (\frac {x}{2}\right )+\left (4 x^2-4 x^2 \log \left (\frac {x}{2}\right )\right ) \log (x)+\left (-e^x+\left (e^x (1-x)-4 x\right ) \log \left (\frac {x}{2}\right )+\left (-4 x+4 x \log \left (\frac {x}{2}\right )\right ) \log (x)\right ) \log (\log (\log (4)))+\left (\log \left (\frac {x}{2}\right )+\left (1-\log \left (\frac {x}{2}\right )\right ) \log (x)\right ) \log ^2(\log (\log (4)))}{x^2 \left (4 x^2-4 x \log (\log (\log (4)))+\log ^2(\log (\log (4)))\right )} \, dx\\ &=\int \frac {2 e^x x+\left (4 x^2+e^x \left (-4 x+2 x^2\right )\right ) \log \left (\frac {x}{2}\right )+\left (4 x^2-4 x^2 \log \left (\frac {x}{2}\right )\right ) \log (x)+\left (-e^x+\left (e^x (1-x)-4 x\right ) \log \left (\frac {x}{2}\right )+\left (-4 x+4 x \log \left (\frac {x}{2}\right )\right ) \log (x)\right ) \log (\log (\log (4)))+\left (\log \left (\frac {x}{2}\right )+\left (1-\log \left (\frac {x}{2}\right )\right ) \log (x)\right ) \log ^2(\log (\log (4)))}{x^2 (2 x-\log (\log (\log (4))))^2} \, dx\\ &=\int \frac {\left (e^x+\log (x) (2 x-\log (\log (\log (4))))\right ) (2 x-\log (\log (\log (4))))+\log \left (\frac {x}{2}\right ) \left ((-2 x+\log (\log (\log (4))))^2-\log (x) (-2 x+\log (\log (\log (4))))^2+e^x \left (2 x^2+\log (\log (\log (4)))-x (4+\log (\log (\log (4))))\right )\right )}{x^2 (2 x-\log (\log (\log (4))))^2} \, dx\\ &=\int \left (\frac {\log \left (\frac {x}{2}\right )+\log (x)-\log \left (\frac {x}{2}\right ) \log (x)}{x^2}+\frac {e^x \left (2 x+2 x^2 \log \left (\frac {x}{2}\right )-4 x \log \left (\frac {x}{2}\right ) \left (1+\frac {1}{4} \log (\log (\log (4)))\right )-\log (\log (\log (4)))+\log \left (\frac {x}{2}\right ) \log (\log (\log (4)))\right )}{x^2 (2 x-\log (\log (\log (4))))^2}\right ) \, dx\\ &=\int \frac {\log \left (\frac {x}{2}\right )+\log (x)-\log \left (\frac {x}{2}\right ) \log (x)}{x^2} \, dx+\int \frac {e^x \left (2 x+2 x^2 \log \left (\frac {x}{2}\right )-4 x \log \left (\frac {x}{2}\right ) \left (1+\frac {1}{4} \log (\log (\log (4)))\right )-\log (\log (\log (4)))+\log \left (\frac {x}{2}\right ) \log (\log (\log (4)))\right )}{x^2 (2 x-\log (\log (\log (4))))^2} \, dx\\ &=\int \left (\frac {\log \left (\frac {x}{2}\right )}{x^2}-\frac {\left (-1+\log \left (\frac {x}{2}\right )\right ) \log (x)}{x^2}\right ) \, dx+\int \frac {e^x \left (2 x-\log (\log (\log (4)))+\log \left (\frac {x}{2}\right ) \left (2 x^2+\log (\log (\log (4)))-x (4+\log (\log (\log (4))))\right )\right )}{x^2 (2 x-\log (\log (\log (4))))^2} \, dx\\ &=\int \frac {\log \left (\frac {x}{2}\right )}{x^2} \, dx-\int \frac {\left (-1+\log \left (\frac {x}{2}\right )\right ) \log (x)}{x^2} \, dx+\int \left (\frac {e^x}{x^2 (2 x-\log (\log (\log (4))))}+\frac {e^x \log \left (\frac {x}{2}\right ) \left (2 x^2+\log (\log (\log (4)))-x (4+\log (\log (\log (4))))\right )}{x^2 (2 x-\log (\log (\log (4))))^2}\right ) \, dx\\ &=-\frac {1}{x}-\frac {\log \left (\frac {x}{2}\right )}{x}+\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\int \frac {\log \left (\frac {x}{2}\right )}{x^2} \, dx+\int \frac {e^x}{x^2 (2 x-\log (\log (\log (4))))} \, dx+\int \frac {e^x \log \left (\frac {x}{2}\right ) \left (2 x^2+\log (\log (\log (4)))-x (4+\log (\log (\log (4))))\right )}{x^2 (2 x-\log (\log (\log (4))))^2} \, dx\\ &=\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{(2 x-\log (\log (\log (4)))) \log (\log (\log (4)))}+\int \left (-\frac {2 e^x}{x \log ^2(\log (\log (4)))}+\frac {4 e^x}{(2 x-\log (\log (\log (4)))) \log ^2(\log (\log (4)))}-\frac {e^x}{x^2 \log (\log (\log (4)))}\right ) \, dx-\int \frac {e^x}{x^2 (2 x-\log (\log (\log (4))))} \, dx\\ &=\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{(2 x-\log (\log (\log (4)))) \log (\log (\log (4)))}-\frac {2 \int \frac {e^x}{x} \, dx}{\log ^2(\log (\log (4)))}+\frac {4 \int \frac {e^x}{2 x-\log (\log (\log (4)))} \, dx}{\log ^2(\log (\log (4)))}-\frac {\int \frac {e^x}{x^2} \, dx}{\log (\log (\log (4)))}-\int \left (-\frac {2 e^x}{x \log ^2(\log (\log (4)))}+\frac {4 e^x}{(2 x-\log (\log (\log (4)))) \log ^2(\log (\log (4)))}-\frac {e^x}{x^2 \log (\log (\log (4)))}\right ) \, dx\\ &=\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\frac {2 \text {Ei}(x)}{\log ^2(\log (\log (4)))}+\frac {2 \text {Ei}\left (\frac {1}{2} (2 x-\log (\log (\log (4))))\right ) \sqrt {\log (\log (4))}}{\log ^2(\log (\log (4)))}+\frac {e^x}{x \log (\log (\log (4)))}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{(2 x-\log (\log (\log (4)))) \log (\log (\log (4)))}+\frac {2 \int \frac {e^x}{x} \, dx}{\log ^2(\log (\log (4)))}-\frac {4 \int \frac {e^x}{2 x-\log (\log (\log (4)))} \, dx}{\log ^2(\log (\log (4)))}+\frac {\int \frac {e^x}{x^2} \, dx}{\log (\log (\log (4)))}-\frac {\int \frac {e^x}{x} \, dx}{\log (\log (\log (4)))}\\ &=\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\frac {\text {Ei}(x)}{\log (\log (\log (4)))}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{(2 x-\log (\log (\log (4)))) \log (\log (\log (4)))}+\frac {\int \frac {e^x}{x} \, dx}{\log (\log (\log (4)))}\\ &=\frac {\log \left (\frac {x}{2}\right ) \log (x)}{x}-\frac {e^x \log \left (\frac {x}{2}\right )}{x \log (\log (\log (4)))}+\frac {2 e^x \log \left (\frac {x}{2}\right )}{(2 x-\log (\log (\log (4)))) \log (\log (\log (4)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 29, normalized size = 1.00 \begin {gather*} \frac {\log \left (\frac {x}{2}\right ) \left (\log (x)+\frac {e^x}{2 x-\log (\log (\log (4)))}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^x*x + (4*x^2 + E^x*(-4*x + 2*x^2))*Log[x/2] + (4*x^2 - 4*x^2*Log[x/2])*Log[x] + (-E^x + (E^x*(1
- x) - 4*x)*Log[x/2] + (-4*x + 4*x*Log[x/2])*Log[x])*Log[Log[Log[4]]] + (Log[x/2] + (1 - Log[x/2])*Log[x])*Lo
g[Log[Log[4]]]^2)/(4*x^4 - 4*x^3*Log[Log[Log[4]]] + x^2*Log[Log[Log[4]]]^2),x]

[Out]

(Log[x/2]*(Log[x] + E^x/(2*x - Log[Log[Log[4]]])))/x

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fricas [B]  time = 0.87, size = 63, normalized size = 2.17 \begin {gather*} \frac {2 \, x \log \left (\frac {1}{2} \, x\right )^{2} + {\left (2 \, x \log \relax (2) + e^{x}\right )} \log \left (\frac {1}{2} \, x\right ) - {\left (\log \relax (2) \log \left (\frac {1}{2} \, x\right ) + \log \left (\frac {1}{2} \, x\right )^{2}\right )} \log \left (\log \left (2 \, \log \relax (2)\right )\right )}{2 \, x^{2} - x \log \left (\log \left (2 \, \log \relax (2)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(1/2*x)+1)*log(x)+log(1/2*x))*log(log(2*log(2)))^2+((4*x*log(1/2*x)-4*x)*log(x)+((-x+1)*exp(x
)-4*x)*log(1/2*x)-exp(x))*log(log(2*log(2)))+(-4*x^2*log(1/2*x)+4*x^2)*log(x)+((2*x^2-4*x)*exp(x)+4*x^2)*log(1
/2*x)+2*exp(x)*x)/(x^2*log(log(2*log(2)))^2-4*x^3*log(log(2*log(2)))+4*x^4),x, algorithm="fricas")

[Out]

(2*x*log(1/2*x)^2 + (2*x*log(2) + e^x)*log(1/2*x) - (log(2)*log(1/2*x) + log(1/2*x)^2)*log(log(2*log(2))))/(2*
x^2 - x*log(log(2*log(2))))

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giac [B]  time = 0.15, size = 71, normalized size = 2.45 \begin {gather*} -\frac {2 \, x \log \relax (2) \log \relax (x) - 2 \, x \log \relax (x)^{2} - \log \relax (2) \log \relax (x) \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right ) + \log \relax (x)^{2} \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right ) + e^{x} \log \relax (2) - e^{x} \log \relax (x)}{2 \, x^{2} - x \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(1/2*x)+1)*log(x)+log(1/2*x))*log(log(2*log(2)))^2+((4*x*log(1/2*x)-4*x)*log(x)+((-x+1)*exp(x
)-4*x)*log(1/2*x)-exp(x))*log(log(2*log(2)))+(-4*x^2*log(1/2*x)+4*x^2)*log(x)+((2*x^2-4*x)*exp(x)+4*x^2)*log(1
/2*x)+2*exp(x)*x)/(x^2*log(log(2*log(2)))^2-4*x^3*log(log(2*log(2)))+4*x^4),x, algorithm="giac")

[Out]

-(2*x*log(2)*log(x) - 2*x*log(x)^2 - log(2)*log(x)*log(log(2) + log(log(2))) + log(x)^2*log(log(2) + log(log(2
))) + e^x*log(2) - e^x*log(x))/(2*x^2 - x*log(log(2) + log(log(2))))

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maple [A]  time = 0.67, size = 47, normalized size = 1.62

method	result	size
default	\(\frac {{\mathrm e}^{x} \ln \relax (2)-{\mathrm e}^{x} \ln \relax (x )}{x \left (\ln \left (\ln \left (2 \ln \relax (2)\right )\right )-2 x \right )}-\frac {\ln \relax (2) \ln \relax (x )}{x}+\frac {\ln \relax (x )^{2}}{x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-ln(1/2*x)+1)*ln(x)+ln(1/2*x))*ln(ln(2*ln(2)))^2+((4*x*ln(1/2*x)-4*x)*ln(x)+((1-x)*exp(x)-4*x)*ln(1/2*x
)-exp(x))*ln(ln(2*ln(2)))+(-4*x^2*ln(1/2*x)+4*x^2)*ln(x)+((2*x^2-4*x)*exp(x)+4*x^2)*ln(1/2*x)+2*exp(x)*x)/(x^2
*ln(ln(2*ln(2)))^2-4*x^3*ln(ln(2*ln(2)))+4*x^4),x,method=_RETURNVERBOSE)

[Out]

(exp(x)*ln(2)-exp(x)*ln(x))/x/(ln(ln(2*ln(2)))-2*x)-ln(2)/x*ln(x)+ln(x)^2/x

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maxima [B]  time = 0.52, size = 70, normalized size = 2.41 \begin {gather*} \frac {{\left (2 \, x - \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )\right )} \log \relax (x)^{2} - {\left (\log \relax (2) - \log \relax (x)\right )} e^{x} - {\left (2 \, x \log \relax (2) - \log \relax (2) \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )\right )} \log \relax (x)}{2 \, x^{2} - x \log \left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-log(1/2*x)+1)*log(x)+log(1/2*x))*log(log(2*log(2)))^2+((4*x*log(1/2*x)-4*x)*log(x)+((-x+1)*exp(x
)-4*x)*log(1/2*x)-exp(x))*log(log(2*log(2)))+(-4*x^2*log(1/2*x)+4*x^2)*log(x)+((2*x^2-4*x)*exp(x)+4*x^2)*log(1
/2*x)+2*exp(x)*x)/(x^2*log(log(2*log(2)))^2-4*x^3*log(log(2*log(2)))+4*x^4),x, algorithm="maxima")

[Out]

((2*x - log(log(2) + log(log(2))))*log(x)^2 - (log(2) - log(x))*e^x - (2*x*log(2) - log(2)*log(log(2) + log(lo
g(2))))*log(x))/(2*x^2 - x*log(log(2) + log(log(2))))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {\ln \relax (x)\,\left (4\,x^2\,\ln \left (\frac {x}{2}\right )-4\,x^2\right )-2\,x\,{\mathrm {e}}^x+\ln \left (\ln \left (2\,\ln \relax (2)\right )\right )\,\left ({\mathrm {e}}^x+\ln \relax (x)\,\left (4\,x-4\,x\,\ln \left (\frac {x}{2}\right )\right )+\ln \left (\frac {x}{2}\right )\,\left (4\,x+{\mathrm {e}}^x\,\left (x-1\right )\right )\right )+\ln \left (\frac {x}{2}\right )\,\left ({\mathrm {e}}^x\,\left (4\,x-2\,x^2\right )-4\,x^2\right )-{\ln \left (\ln \left (2\,\ln \relax (2)\right )\right )}^2\,\left (\ln \left (\frac {x}{2}\right )-\ln \relax (x)\,\left (\ln \left (\frac {x}{2}\right )-1\right )\right )}{4\,x^4-4\,\ln \left (\ln \left (2\,\ln \relax (2)\right )\right )\,x^3+{\ln \left (\ln \left (2\,\ln \relax (2)\right )\right )}^2\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(4*x^2*log(x/2) - 4*x^2) - 2*x*exp(x) + log(log(2*log(2)))*(exp(x) + log(x)*(4*x - 4*x*log(x/2))
+ log(x/2)*(4*x + exp(x)*(x - 1))) + log(x/2)*(exp(x)*(4*x - 2*x^2) - 4*x^2) - log(log(2*log(2)))^2*(log(x/2)
- log(x)*(log(x/2) - 1)))/(x^2*log(log(2*log(2)))^2 + 4*x^4 - 4*x^3*log(log(2*log(2)))),x)

[Out]

-int((log(x)*(4*x^2*log(x/2) - 4*x^2) - 2*x*exp(x) + log(log(2*log(2)))*(exp(x) + log(x)*(4*x - 4*x*log(x/2))
+ log(x/2)*(4*x + exp(x)*(x - 1))) + log(x/2)*(exp(x)*(4*x - 2*x^2) - 4*x^2) - log(log(2*log(2)))^2*(log(x/2)
- log(x)*(log(x/2) - 1)))/(x^2*log(log(2*log(2)))^2 + 4*x^4 - 4*x^3*log(log(2*log(2)))), x)

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sympy [A]  time = 0.53, size = 41, normalized size = 1.41 \begin {gather*} \frac {\left (\log {\relax (x )} - \log {\relax (2 )}\right ) e^{x}}{2 x^{2} - x \log {\left (\log {\left (\log {\relax (2 )} \right )} + \log {\relax (2 )} \right )}} + \frac {\log {\relax (x )}^{2}}{x} - \frac {\log {\relax (2 )} \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-ln(1/2*x)+1)*ln(x)+ln(1/2*x))*ln(ln(2*ln(2)))**2+((4*x*ln(1/2*x)-4*x)*ln(x)+((-x+1)*exp(x)-4*x)*
ln(1/2*x)-exp(x))*ln(ln(2*ln(2)))+(-4*x**2*ln(1/2*x)+4*x**2)*ln(x)+((2*x**2-4*x)*exp(x)+4*x**2)*ln(1/2*x)+2*ex
p(x)*x)/(x**2*ln(ln(2*ln(2)))**2-4*x**3*ln(ln(2*ln(2)))+4*x**4),x)

[Out]

(log(x) - log(2))*exp(x)/(2*x**2 - x*log(log(log(2)) + log(2))) + log(x)**2/x - log(2)*log(x)/x

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